Math 210A Lecture 10 Notes

Daniel Raban

October 19, 2018

1 Images, Coimages, and Generating Sets

1.1 Images Definition 1.1. The image im(f) of f : A → B is an object and a monomorphism ι : im(f) → B such that there exists π : A → im(f) with π ◦ ι and such that if e : C → B is a monomorphism and g : A → C is such that e ◦ g = f, then there exists a unique morphism ψ : im(f) → C such that g ◦ ψ = ι.

f A B π ι

g im(f) e ψ C

Example 1.1. In Set, f(A) = im(f). Then b ∈ F (A) =⇒ b = f(a) for some a ∈ A. Then g(a) ∈ C is the unique element with e(g(a)) = (a) because e is a monomorphism. So ψ(f(a)) = g(a). Proposition 1.1. If C has equalizers, then π : A → im(f) is an epimorphism. Proof. Suppose α A ι im(f) D β commutes. Then α ◦ π = β ◦ π,

π A eq(α, β) c im(f) f ι B

1 Then there is a unique d : im(f) → eq(α, β), and c ◦ d = id and d ◦ c = id by uniqueness. So (im(f), idim(f)) equalizes α im(f) D β so α = β.

Suppose that in C, every morphism factors through an equalizer and the category has finite limits and colimits. Then im(f) can be defined as the equalizer of the following diagram: ι1 B B qA B ι2 We get the following diagram.

A π f f im(f)

ι ι B B

ι2 ι1

B qA B

1.2 Coimages Definition 1.2. The coimage im(f) of f : A → B is an object and a monomorphism π : A → coim(f) such that there exists ι : coim(f) → B such that ι ◦ π and such that if g : A → C is an epinmorphism and e : C → B is such that e ◦ g = f, then there exists a unique morphism θ : C → coim(f) such that θ ◦ g = π.

f A B π ι

g coim(f) e ψ C

So ι ◦ θ ◦ g = ι ◦ π = f = e ◦ g. Since g is an epimorphism, i ◦ θ = e. How are the image and coimage related?

2 f A B

im(f) coim(f)

Definition 1.3. A morphism f : A → B is strict if im(f) → coim(f) is an isomorphism.

In Grp, Ring, Rmod, Set, and Top, im(f) is the set theoretic image. The coimages are quotient objects (of A).

Example 1.2. In Set, coim(f) = A/ ∼, where a ∼ a; if f(a) ∼ f(a0). All the morphisms are strict.

Example 1.3. In Gp, coim(f : C → C0) = G/ ker(f). im(f) ⊆ f(G) ≤ G0. So the image and coimage are isomorphic, which is the first isomorphism theorem.

Example 1.4. In Ring, let ker(f) be the category theoretic kernel. Then coim(f) = R/ ker(f) −→∼ im(f) by the first isomorphism theorem.

Example 1.5. In the category of left R-modules, morphisms are also strict.

1.3 Generating sets Definition 1.4. Let Φ : C → Set be a faithful functor, and let F be a left adjoint to Φ. f Let FX = F (X) be the free object on X. If X −→ Φ(A) for A ∈ C, we get φ : FX → A. Suppose im(φ) exists. Then im(φ) is called the subobject of A generated by X.

Example 1.6. In Gp, let X ⊆ G. Then hXi is the subgroup of G generated by X. This n1 xr n1 nr n1 nr is im(φ : TX → G), where φ(x1 ··· xr ) = x1 ··· xr . So this is {x1 ··· xr : x1 ∈ X, ni ∈ Z, 1 ≤ i ≤ r, r ≥ 0}. We claim that hXi is the smallest subgroup of G containing X, or equivalently, the intersection of all subgroups of G containing X. Indeed, this is a subgroup of G containing X, and any subgroup of G containing X must contain these words, since it must be closed under products. Pn Example 1.7. In Rmod, if X ⊆ A, R · X = { i=1 rixi : ri ∈ R, xi ∈ X, 1 ≤ i ≤ n, n ≥ 0}. L φ So FX = x∈X Rx −→ A, where φ(r · x) = rx ∈ A. Pn Example 1.8. In the category of (R,S)-bimodules, RXS = { i=1 rixisi : ri ∈ R, si ∈ Pn S, xi ∈ X, 1 ≤ i ≤ n, n ≥ 0}. If we have the set of formal sums RxS = { i=1 rixsi : ri ∈ 0 0 0 0 R, si ∈ S, 1 ≤ i ≤ n, n ≥ 0} with (r + r )xs = rxs + r xs and rx(s + s ) = rxs + rxs , then L the free object is x∈X RxS. Ideals uses (R,R)-subbimodules of R generated by X ⊆ R.

3 Pn 0 0 Definition 1.5. The ideal generated by X is (X) = { i=1 rixiri : ri, ri ∈ R, xi ∈ X}. If X = {x1, . . . , xn}, then we write (x1, . . . , xn). Remark 1.1. Even if X = {x}, we still need to take sums to get (x).

4